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If ab ≠ 0, is ab > a+b? [#permalink]
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06 May 2016, 08:46
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If \(ab ≠ 0\), is \(ab > a+b\)? 1) \(ab < 0\) 2) \(a > b\)
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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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06 May 2016, 09:21
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For ab > a+b, a and b should have opposite signs. So, if a > 0, b should be less than 0 and vice versa.
1) ab < 0 this proves a and b have opposite signs. So, sufficient.
2) a>b not sufficient as it does not prove a and b will have opposite signs. for e.g. a=1, b=2 > 3 > 1 > Yes a=3, b =2 > 1 < 5 > No



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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06 May 2016, 09:25
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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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06 May 2016, 09:31
Bunuel wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
Square \(ab > a+b\) (we can safely do this since both sides are nonnegative): is \(a^2  2ab + b^2 > a^2 +2ab + b^2\) > is \(ab < 0\)?
(1) \(ab < 0\). Directly answers the question. Sufficient.
(2) \(a > b\). Not sufficient to say whether ab < 0.
Answer: A. Thanks for the reply, Bunuel! Just a quick followup: can we assume that both sides are nonnegative since ab ≠ 0?



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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06 May 2016, 09:35



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If ab ≠ 0, is ab > a+b? [#permalink]
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Hope this helps
Originally posted by Z10201Z on 06 May 2016, 09:52.
Last edited by Z10201Z on 06 May 2016, 09:54, edited 1 time in total.



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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06 May 2016, 09:53
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Sallyzodiac wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
1) \(ab < 0\)
2) \(a > b\) Hi, Ofcourse the above method by Bunuel is the right and best method, But what it means logically.. is \(ab > a+b\)?.. We have a number line in which a and b and b are marked.. the MOD asks us whether a is farther from b than b.. two cases A) if both are same sign, a will be closer to b than b, since midpoint of b and b will be 0.. ans will be NO B) if both are opposite sign, a will be closer to b and ans will be YES..
1) \(ab < 0\) Both a and b are of opposite sign .. ans YESSuff 2) \(a > b\) Nothing .. both can be same sign or opposite sign .. the above TWO cases still remain.. Insuff
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If ab ≠ 0, is ab > a+b? [#permalink]
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07 May 2016, 12:36
chetan2u wrote: Sallyzodiac wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
1) \(ab < 0\)
2) \(a > b\) Hi, Ofcourse the above method by Bunuel is the right and best method, But what it means logically.. is \(ab > a+b\)?.. We have a number line in which a and b and b are marked.. the MOD asks us whether a is farther from b than b.. two cases A) if both are same sign, a will be closer to b than b, since midpoint of b and b will be 0.. ans will be NO B) if both are opposite sign, a will be closer to b and ans will be YES..
1) \(ab < 0\) Both a and b are of opposite sign .. ans YESSuff 2) \(a > b\) Nothing .. both can be same sign or opposite sign .. the above TWO cases still remain.. Insuff Hi chetan2u, I tried other approach by opening the modulus but it seems there is an error. is a−b>a+b? case 1: ab > a+b……… 2b<0……..so b<0.. B is negative case 2: ab < ab …….. 2a<0……..so a< 0 ..a is negative Statement 1: ab<0 According to my analysis above, it should be insufficient because both a & b should negative. Statement 2: a>b It should be insufficient Can you help with identifying the problem with above solution? Thanks in advance



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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07 May 2016, 16:04
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Mo2men wrote: chetan2u wrote: Sallyzodiac wrote: If \(ab ≠ 0\), is \(ab > a+b\)?
1) \(ab < 0\)
2) \(a > b\) Hi, Ofcourse the above method by Bunuel is the right and best method, But what it means logically.. is \(ab > a+b\)?.. We have a number line in which a and b and b are marked.. the MOD asks us whether a is farther from b than b.. two cases A) if both are same sign, a will be closer to b than b, since midpoint of b and b will be 0.. ans will be NO B) if both are opposite sign, a will be closer to b and ans will be YES..
1) \(ab < 0\) Both a and b are of opposite sign .. ans YESSuff 2) \(a > b\) Nothing .. both can be same sign or opposite sign .. the above TWO cases still remain.. Insuff Hi chetan2u, I tried other approach by opening the modulus but it seems there is an error. is a−b>a+b? case 1: ab > a+b……… 2b<0……..so b<0.. B is negative case 2: ab < ab …….. 2a<0……..so a< 0 ..a is negative Statement 1: ab<0 According to my analysis above, it should be insufficient because both a & b should negative. Statement 2: a>b It should be insufficient Can you help with identifying the problem with above solution? Thanks in advance Take a look at my previous answer. You must analyze four different situations because the two sides of the equation are not independent. Hope it helps Best regards



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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07 May 2016, 19:23
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Mo2men wrote: Hi chetan2u,
I tried other approach by opening the modulus but it seems there is an error.
is a−b>a+b? case 1: ab > a+b……… 2b<0……..so b<0.. B is negative case 2: ab < ab …….. 2a<0……..so a< 0 ..a is negative Statement 1: ab<0 According to my analysis above, it should be insufficient because both a & b should negative. Statement 2: a>b It should be insufficient
Can you help with identifying the problem with above solution?
Thanks in advance
Hi, the problem in your solution is opening of MOD.. yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it.. Do not waste time in taking four scenarios, but square both sides, it will siffice.. \(ab>a+b..\) square both sides.. \((ab)^2 > (a+b)^2................... a^2+b^22ab>a^2+b^2+2ab............ 4ab<0....\).... so we should find out if ab<0 and statement I tells us that ab<0 .. so suff
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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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07 May 2016, 23:31
chetan2u wrote: Mo2men wrote: Hi chetan2u,
I tried other approach by opening the modulus but it seems there is an error.
is a−b>a+b? case 1: ab > a+b……… 2b<0……..so b<0.. B is negative case 2: ab < ab …….. 2a<0……..so a< 0 ..a is negative Statement 1: ab<0 According to my analysis above, it should be insufficient because both a & b should negative. Statement 2: a>b It should be insufficient
Can you help with identifying the problem with above solution?
Thanks in advance
Hi, the problem in your solution is opening of MOD.. yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it.. Do not waste time in taking four scenarios, but square both sides, it will siffice.. \(ab>a+b..\) square both sides.. \((ab)^2 > (a+b)^2................... a^2+b^22ab>a^2+b^2+2ab............ 4ab<0....\).... so we should find out if ab<0 and statement I tells us that ab<0 .. so suff Thanks Z10201Z for your help Hi chetan2u, Thanks you for your quick response. I certainly will follow the rule in case of two modulus with inequality. Aster reading your post about 'Absolute modulus : A better understanding', I have two questions: 1 I want to take further to solve a confusion. I have seen your way solving in one problem when two modulus in equality: X=Y X=Y or X=Y or X=Y (which is same as X=Y) or X=Y (Which is the same as X=Y). However, the previous technique is valid only for Equity and invalid for inequality. Am I right? I think this my error. 2 if the question is ab>a+b ............what the best to solve it? In the case, I'm curious to know how to open modulus just to get to know. I hope one day you can make another post for absolute modulus under inequality. Thanks in advance for your help. I appreciate it.



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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08 May 2016, 23:05
Mo2men wrote: chetan2u wrote: Mo2men wrote: Hi chetan2u,
I tried other approach by opening the modulus but it seems there is an error.
is a−b>a+b? case 1: ab > a+b……… 2b<0……..so b<0.. B is negative case 2: ab < ab …….. 2a<0……..so a< 0 ..a is negative Statement 1: ab<0 According to my analysis above, it should be insufficient because both a & b should negative. Statement 2: a>b It should be insufficient
Can you help with identifying the problem with above solution?
Thanks in advance
Hi, the problem in your solution is opening of MOD.. yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it.. Do not waste time in taking four scenarios, but square both sides, it will siffice.. \(ab>a+b..\) square both sides.. \((ab)^2 > (a+b)^2................... a^2+b^22ab>a^2+b^2+2ab............ 4ab<0....\).... so we should find out if ab<0 and statement I tells us that ab<0 .. so suff Thanks Z10201Z for your help Hi chetan2u, Thanks you for your quick response. I certainly will follow the rule in case of two modulus with inequality. Aster reading your post about 'Absolute modulus : A better understanding', I have two questions: 1 I want to take further to solve a confusion. I have seen your way solving in one problem when two modulus in equality: X=Y X=Y or X=Y or X=Y (which is same as X=Y) or X=Y (Which is the same as X=Y). However, the previous technique is valid only for Equity and invalid for inequality. Am I right? I think this my error. 2 if the question is ab>a+b ............what the best to solve it? In the case, I'm curious to know how to open modulus just to get to know. I hope one day you can make another post for absolute modulus under inequality. Thanks in advance for your help. I appreciate it. hi lets solve ab>a+b.. this can be straight .. 1) ab>a+b......2b<0..... b<0.. 2) (ab)>a+b......a+b>a+b.....2a<0..... a<0.. Or Critical points.. a is 0, when a is b, so b is our critical point.. this gives us two ranges a<b or a>=b.. 1) a<b... then ab will be ive.. so (ab)>a+b......a+b>a+b.....2a<0..... a<0.. 2) a>=b.. then ab will be +ive or 0.. so ab>a+b......2b<0..... b<0.. Best METHOD for any MOD irrespective of = or inequality, follow "critical point" method
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If ab ≠ 0, is ab > a+b? [#permalink]
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08 May 2016, 23:39
chetan2u wrote: hi lets solve ab>a+b.. this can be straight .. 1) ab>a+b......2b<0..... b<0.. 2) (ab)>a+b......a+b>a+b.....2a<0..... a<0..
Or Critical points..
a is 0, when a is b, so b is our critical point.. this gives us two ranges a<b or a>=b.. 1) a<b... then ab will be ive.. so (ab)>a+b......a+b>a+b.....2a<0..... a<0.. 2) a>=b.. then ab will be +ive or 0.. so ab>a+b......2b<0..... b<0..
Best METHOD for any MOD irrespective of = or inequality, follow "critical point" method Thanks a lot for your reply. I think there is typo. it should: ab=0, when a is b



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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15 May 2016, 04:27
Rewording the question , ab > a+b, a and b should have opposite signs.
Statement 1 exactly points out that : Hence sufficient.
Statement 2 Insufficient



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Re: If ab ≠ 0, is ab > a+b ? [#permalink]
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25 Jun 2016, 10:42
Is a  b > a + b ? Square on both sides > Is a^2 + b^2  2ab > a^2 + b^2 + 2ab > Is 4ab < 0 ? Is ab < 0?
St1: ab < 0 > Directly answers the question. Sufficient
St2: a > b > Clearly Insufficient. If a = 2 and b = 1 then ab > 0 If a = 2 and b = 1 then ab < 0
Answer: A



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If ab ≠ 0, is ab > a+b ? [#permalink]
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25 Jun 2016, 11:01
shiv242003 wrote: If ab ≠ 0, is ab > a+b ?
(1) ab < 0 (2) a > b
Could some one please help me with this question with a valid explanation. Thanks , Shiv. According to statement 1. a*b is negative. So, either a or b is negative. Both of them can't be negative. by question a*b doesn't equals to zero. take testing numbers. a = 1 or 1 b = 1 or 1 if a = 1 and b =1 then; 1 (1) > 1 1 or, 1+1 > 0 or, 2>0. true. also, check for a= 1 and b =1 1(1) > 1+1 or, 11 > 0 or, 2>0. true. so Statement 1 is sufficient. According to statement 2, a>b. take testing points. a=1 or 2 b= 2 or 1 if a= 1 and b = 2 then, 1  (2) > 1+(2) or, 1+2> 12 or, 1>3 or 1>3 false. If a= 2 and b = 1 2(1)>21 or,3>1 or, 3>1 true. So, statement 2 is not sufficient to answer the question. Statement 2 gives either true or false. Whereas, Statement 1 gives true only. So, answer A. I was waiting for an easy solution and Vyshak gives that.



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Re: If ab ≠ 0, is ab > a+b ? [#permalink]
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You can think of absolute value as distance from zero. So we need enough info to definitely determine whether a  b is further from zero than the sum a + b.
Statement one tells us that a and b have different signs (that's the only way that the product will be negative).
For the difference a  b, this means that the numbers end up 'working together' to get away from zero; if you take a positive number away from a negative number, you go further down, further away from zero (e.g. 10  2 = 12). If you take a negative number away from a positive number you go further up (e.g. 10   2 = 12) Takeaway: The difference of numbers with opposite signs is always further from zero than either of the original numbers.
For the sum a + b, statement one means that the numbers end up 'working in opposite directions.' That is, when we add numbers of opposite signs, one cancels out part of the other's distance to zero. Think of 10 +  2 and 2 + 10. In both cases we end up closer to zero than the furthest term was to begin with. Takeaway: the sum of two numbers with opposite signs will be closer to zero than the original number of greatest magnitude (the term that was furthest from zero to begin with).
So if we know two numbers have opposite signs, we know the difference of those numbers ends up further from zero than the sum. Sufficient.
statement 2 doesn't tell us anything about the signs, so we might be suspicious right from the start. And sure enough, if all we know is that a is to the right of b, then we might have a couple of positives, (such as a = 3, and b = 2) in which case the sum is further from zero than the difference. But we could also have a positive and a negative (such as a = 3, and b =  2), in which case the difference is further from zero than the sum. Insufficient.



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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26 Oct 2016, 11:33
karant wrote: If ab is not equal to 0, is ab > a+b ?
(1) ab < 0
(2) a > b
Need help on this official question. We can square both sides of 'ab > a+b' and we will get 4ab<0.1. Gives us the ab<0. Bingo. That is what required to get the above expression true. Sufficient. 2. a>b. Clearly not sufficent as A and be could be 3 and 2 respectively or 2 and 3. Insufficient. So answer is A.



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Re: If ab ≠ 0, is ab > a+b? [#permalink]
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Hello Bunuel, chetan2uThanks for the solution. I got the point. However, I would like to know how this problem will be solved algebrically. Can you please explain the solution by opening the modulus on both the side ? There will be following two condition if we open modulus: ab > a+b gives, b <0 and ab > (a+b) gives , a>0 What would be the next steps now ? How to interpret. Please explain. Thanks



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Re: If ab ≠ 0, is ab>a+b? [#permalink]
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07 Jan 2017, 08:35
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nayanparikh wrote: If ab ≠ 0, is ab>a+b?
1) ab < 0 2) a > b
Can any one please provide explanation on how to solve these type of questions ? @Bunnel ? 1) ab < 0 Means that there are two possibilities: either a>0 and b<0 or a<0 and b>0. Let's look at htose two seperately. a>0, b<0. Let's denote b=y (y  some positive integer). y<0, y>0. (a  (y)) > a  y a + y > a  y On the left side we have sum of two positive integers so we can pull off our modulus. On the right side we have difference o two positive integers which will always be lesser than their sum. a<0, b>0. Same approach. a=x (x some positive integer). x<0, x>0 x  b > x + b   (x +b) > b  x x + b > b  x Again we have sum of two positive integers on the left and their diffrence on th right. Sufficient. 2) a > b We have three possible scenarious. a) a>0, b>0 b) a<0, b<0 c) a>0, b<0 a) To eveade confusion it would be impler to use particular numbers here. a = 30, b = 10 30  10 ... 30 + 10 > 20 < 40 b) a = 10 b = 30 10  (30) ...  10  30  10 + 30 ...   (10 + 30) 20 < 40 c) In c we have same case as in (1) a=10, b=30 10  (30) ... 10  30 40 > 20 Insufficient Answer A I hope my approach wasn't too vague. May be our experts will give you more details.




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